Activation energy `(E_a)` and rate constants `(k_1 and k_2)` of a chemical reaction at two different temperature `(T_1 and T_2)` are related by

Options

A. In `k_2/k_1=-E_a/R(1/T_1-1/T_2)`
B. In `k_2/k_1=-E_a/R(1/T_2-1/T_1)`
C. In `k_2/k_1=-E_a/R(1/T_2+1/T_1)`
D. In `k_2/k_1=E_a/R(1/T_1-1/T_2)`

Answer

To derive the relationship between the activation energy `(E_a)` and the rate constants `(k_1 and k_2)` of a chemical reaction at two different temperatures `(T_1 and T_2)`, we will use the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation relates the rate constant `k` to the temperature `T` and the activation energy `E_a`:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = Arrhenius constant (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Write the Equations for Two Different Temperatures
For two different temperatures \( T_1 \) and \( T_2 \), we can write:
\[ k_1 = A e^{-\frac{E_a}{RT_1}} \quad \text{(1)} \]
\[ k_2 = A e^{-\frac{E_a}{RT_2}} \quad \text{(2)} \]

### Step 3: Take Natural Logarithm of Both Equations
Taking the natural logarithm of both equations, we have:
\[ \ln k_1 = \ln A - \frac{E_a}{RT_1} \quad \text{(3)} \]
\[ \ln k_2 = \ln A - \frac{E_a}{RT_2} \quad \text{(4)} \]

### Step 4: Subtract the Two Equations
Now, we subtract equation (4) from equation (3):
\[ \ln k_1 - \ln k_2 = \left(\ln A - \frac{E_a}{RT_1}\right) - \left(\ln A - \frac{E_a}{RT_2}\right) \]
This simplifies to:
\[ \ln k_1 - \ln k_2 = -\frac{E_a}{RT_1} + \frac{E_a}{RT_2} \]

### Step 5: Simplify the Expression
Rearranging gives:
\[ \ln k_1 - \ln k_2 = E_a \left(\frac{1}{RT_2} - \frac{1}{RT_1}\right) \]
Using the property of logarithms, we can express this as:
\[ \ln \left(\frac{k_1}{k_2}\right) = E_a \left(\frac{1}{RT_2} - \frac{1}{RT_1}\right) \]

### Step 6: Final Relation
Thus, we arrive at the final relation:
\[ \ln \left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]